A survey in a magazine that asks readers to return the survey would result in:

Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

good morning you are you still

8 0

3 years ago

Which of the following is not a benefit of increased energy efficiency?

Rzqust [24]

From what I can see it's D, I did this by simply examining the other answers and seeing that they are beneficial, so, from that information, this one must not be.

7 0

3 years ago

When drinking at a private even, you should assume that drinks will ___.

allochka39001 [22]

When drinking at a private event, you should assume that drinks will be STRONGER THAN NORMAL.

At private events, some hosts have the habit of mixing different drinks together in order to increase the intoxicating power of the drinks. This does not normally happen when one is buying from restaurants or other commercial places. Thus, to be on the safe side, one should always assume that drinks will be stronger when one is attending a private event, this will caution one to drink responsibly in order to avoid intoxication.

8 0

3 years ago

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

sergeinik [125]

Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²

5 0

3 years ago

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0

3 years ago